Right triangle, perimeter = area?

Euclid's formula for Pythagorean Triples states that the sides of a right angled triangle must be in the form of $m^2-n^2$ , $2mn$ and $m^2+n^2$ , where $m,n$ are both postive integers. Hence we can now equate the area and perimeter:

$\frac{1}{2}(2mn)(m^2-n^2) = (m^2+n^2)+(m^2-n^2)+(2mn)$

$mn(m^2-n^2)=2m^2+2mn$

$mn(m+n)(m-n)=2m(m+n)$

$mn(m-n)=2m$

$n(m-n)=2$

From this we obtain the $2$ solutions for $(m,n)$ which are $(3,1)$ and $(3,2)$ since $2$ only has $2$ factors. Substituting these values back into the side lengths of the triangle, we obtain the $6-8-10$ triangle with area and perimeter of $\color{#3D99F6}24$ and the $5-12-13$ triangle with area and perimeter $\color{#D61F06}30$ .

Hence $\color{#3D99F6}24+\color{#D61F06}30 = \color{#333333}\boxed{54}$

If $a$ , $b$ , and $\sqrt{a^2 + b^2}$ are the sides of a right triangle, then $P = a + b + \sqrt{a^2 + b^2}$ and $A = \displaystyle \frac {ab}{2}$ . Equating both expressions yields

$\displaystyle a^2 + b^2 = \frac{a^2b^2}{4} - ab^2 - a^2b + 2ab + a^2 + b^2$

which simplifies to

$\displaystyle \begin{aligned} a + b = \frac{ab}{4} + 2 \tag{1} \end{aligned}$

Since $a, b \in \mathbb{N}$ , it follows that $\displaystyle \frac{ab}{4} \in \mathbb{N}$ , and $4$ divides $a$ or $b$ , or $2$ divides both $a$ and $b$ . In the latter case, letting $a = 2p$ and $b = 2q$ , we find that $2p + 2q = pq + 2$ . Therefore, $2$ divides $p$ or $q$ , and so either $a$ or $b$ is divisible by $4$ , which was the former case.

Suppose that $b = 4m$ for some positive integer $m$ . Substituting $(1)$ , we get $a + 4m = am + 2$ . Since this can be written as

$(a - 4) + 2 = (a - 4)m$

we see that $(a - 4) \mid 2$ . Thus, the only solutions to our problem occur for $a = 5$ and $a = 6$ , with respective values $b = 12$ and $b = 8$ . Their corresponding areas are

$\displaystyle \begin{aligned} A & = \frac 12 \cdot 5 \cdot 12 = 30 \\ A & = \frac 12 \cdot 6 \cdot 8 = 24 \end{aligned}$

Our answer is $\boxed{54}$ .

Suppose the 3 sides of the right angled triangle is $a, b$ and $c$ , where $a\le b \le c$ . Then

$\large \begin{cases} a+b+c=p\\ \frac{1}{2}ab=p\\ a^2+b^2=c^2. \end{cases}$

The first equation gives $a+b= p-c$ , square both sides of the equation to obtain $a^2+b^2+2ab=p^2-2pc+c^2.$

The above equation can be simplified using the second and third equations: $4p = p^2 - 2pc .$

This implies that $c=\frac{p-4}{2}$ . Thus $a+b =p-\frac{p-4}{2} =\frac{p+4}{2}$ .

From $a+b =\frac{p+4}{2}$ and $ab=2p$ , we know that $a$ and $b$ are roots of $x^2-\frac{(p+4)x}{2}+2p=0$ .

Since $a$ and $b$ are integers, the discriminant is a perfect square, i.e. $\left(\frac{p+4}{2}\right) ^2 -4(1)(2p)=k^2$ , where $k >0$ is an integer. This equation is equivalent to $p^2-24p+16 =(2k)^2$ , which can be written as $(p-12+2k)(p-12-2k)=128 = 2^7$ . As $p-12+2k$ and $p-12-2k$ has the same parity, and $p-12+2k > p-12-2k$ , we have the following possibles cases.

$\large \begin{cases} p-12+2k=64 \, \, \,\text{and} \, \, \, p-12-2k =2\\ p-12+2k=32 \, \, \,\text{and} \, \, \, p-12-2k =4\\ p-12+2k=16 \, \, \,\text{and} \, \, \, p-12-2k =8. \end{cases}$

The first case give non-integer $k$ , and it is rejected. Solving for the second and the third cases give $p=30$ and $p=24$ respectively.

Hence the sum of all such $p$ is $\boxed{54}$ .

Check: if $p=30$ , $(a,b,c)=(5,12,13)$ ; if $p=24$ , $(a,b,c)=(6,8,10)$ .

Assume the legs are $x$ and $y$ . W.l.o.g, $x \geq y$ .

Perimeter $= x+y+\sqrt{x^2 +y^2}$

Area $=\frac{xy}{2}$

$\therefore x+y+\sqrt{x^2 +y^2}=\frac{xy}{2}$

After squaring and rearranging the equation,

$xy -4x-4y+8=0$

$(x-4)(y-4)=8=8\times1=4\times2$

So the solutions are $(x,y)=(12,5),(6,8)$

Both solutions are accepted since $\sqrt{x^2 +y^2}$ are positive integers. Namely 13 and 10 respectively.

$W=\frac{xy}{2} = 30 \text{ or } 24$

$\therefore \sum{W} = \boxed{54}$